Hence for any \( k \in \mathbb {N}\)  \( \Vert P_{k,n}x \Vert _n = \Vert x\Vert _n\)  and \( f_n(P_{k,n}x) = \Vert x\Vert _n.\)  Hence for any \( k \in \mathbb {N}\)  \( \frac{P_{k,n}x}{\Vert P_{k,n}x \Vert _n}\)  is a norming point for \(f_n.\)  Since \((X \Vert  \cdot \Vert _n)\)  is strictly convex,
\(f_n\)  has exactly one norming point with respect to \( \Vert  \cdot \Vert _n.\)  Hence for any \( k \in \mathbb {N},\)  \( P_{k,n}x = x,\)  which shows that \( x \in F.\)  By ([1]}, Theorem 2 and Lemma 3), \(F = Fix(P_n)\)  is a contractive subset of \((X, \Vert \cdot \Vert _n).\) 
Let \( Q_n:X \rightarrow F\)  be a contractive mapping with respect to \( \Vert  \cdot \Vert _n.\)  Define for any \( M \ge 1 \)
\(N_M(F) = \lbrace  f:X \rightarrow X: \Vert  y - f(x) \Vert  \le M \Vert y-x\Vert  \hbox{ for any } y\in F, x \in X \rbrace .\) 
